The potential due to an electrostatic charge | vedclass

Name: PDF Generator App | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

$(a)$ Electric field due to given charge distribution is

$E=-\frac{d V}{d r}=-\frac{d}{d r} \frac{q e^{-\alpha r}}{4 \pi \varepsilon_{0} r}$

$=\

Vedclass · Accepted Answer

$2q / 1 e$

Vedclass · Answer

$(a)$ Electric field due to given charge distribution is

$E=-\frac{d V}{d r}=-\frac{d}{d r} \frac{q e^{-\alpha r}}{4 \pi \varepsilon_{0} r}$

$=\frac{-q}{4 \pi \varepsilon_{0}} \frac{d}{d r}\left(\frac{e^{-\alpha r}}{r}ight)$

$=\frac{-q}{4 \pi \varepsilon_{0}}\left(\frac{-\alpha r e^{-\alpha \alpha_{r}}-e^{-\alpha r}}{r^{2}}ight)$

$=\frac{q}{4 \pi \varepsilon_{0}} \cdot e^{-\alpha r} \cdot\left(\frac{\alpha r+1}{r^{2}}ight)$

Elcctric field at $r=\frac{1}{1}$ is

Electric field at $r=\frac{1}{\alpha}$ is

$E\left(r=\frac{1}{\alpha}ight)=\frac{q}{4 \pi \varepsilon_{0}} \cdot e^{-\alpha 	imes \frac{1}{\alpha}}\left(\frac{\alpha 	imes \frac{1}{\alpha}+1}{1 / \alpha^{2}}ight)$

$=\frac{(q / e)}{4 \pi \varepsilon_{0}} \cdot 2 \alpha^{2}$

Flux through a sphere of radius $\frac{1}{\alpha}$ is

$\phi=\int E \cdot d A$

For spherical distribution,

$E. d A =E d A \cos 0^{\circ}=E d A$

and $E=$ uniform

So, we have

$	herefore \quad \phi=E \int d A$

$=\frac{ q / e}{4 \pi \varepsilon_{0}} \cdot 2 \alpha^{2} \cdot 4 \pi\left(\frac{1}{\alpha}ight)^{2}$

or $\quad \phi=\frac{2 q}{\varepsilon_{0} e}$

From Gauss' law, we have $\phi=\frac{q_{	ext {enclosed }}}{\varepsilon_{0}}$

So, charge enclosed in given sphere is $\frac{2 q}{e}$.

The potential due to an electrostatic charge distribution is $V(r)=\frac{q e^{-\alpha e r}}{4 \pi \varepsilon_{0} r}$, where $\alpha$ is positive. The net charge within a sphere centred at the origin and of radius $1/ \alpha$ is

Similar Questions

If the electric potential at any point $(x, y, z) \,m$ in space is given by $V =3 x ^{2}$ volt. The electric field at the point $(1,0,3) \,m$ will be ............

Figure shows two equipotential lines in $x, y$ plane for an electric field. The scales are marked. The $x-$ component $E_x$ and $y$ -component $E_y$ of the electric field in the space between these equipotential lines are respectively :-

In a certain region of space, the potential is given by : $V = k[2x^2 - y^2 + z^2].$ The electric field at the point $(1, 1, 1) $ has magnitude =

In which region magnitude of $x$ -component of electric field is maximum, if potential $(V)$ versus distance $(X)$, graph is as shown?

The electric potential $V(x)$ in a region around the origin is given by $V(x) = 4x^2\,volts$ . The electric charge enclosed in a cube of $1\,m$ side with its centre at the origin is (in coulomb)